Bullet fired through 8.1cm board at 452 m/s

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another one!

An indestructible bullet 1.86cm long is fired straight through a board that is 8.1cm thick. the bullet strikes the board with a speed of 452 m/s and emerges with a speed of 298 m/s. what is the average acceleration of the bullet through the board? answer in units of m/s^2

what would the length of the bullet and the width of the wall be used for in a kinematic equation? I know 452=v. and 298=v ... or is that wrong?
 
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Use the same formula for acceleration as your tennis ball question.
 
Think. You have to find the acceleration. List the things you know. Then look at your equations again. List all things that are relavant to kinematics.
 
If you don't know it label it as just t=?. But look at your equations, they all don't have something in it. Each one doesn't have one of these a,v,t,d so which oe should you use.
 
v=v.+a(t)

i know v=452m/s
v.=298 m/s
a= unknown
but i don't know what t should be, and it is known isn't it?
 
Do you need time?
Your missing a variable, distance.
There are other equations you can use.
 
Yes, List out everything you know before you even start a problem. This will help with some headaches. Look for what you don't know also and choose your equation with this information.
 
The bullet is only deaccelerated over 8.1cm. The width of the board.
 
can you explain it in a different way. I don't understand how that fits into the equation.
 
Your initial x=0 because the bullet has not been acted upon yet so your final x=8.1cm
 
ok so the equation is 298^2=452^2+2a(8.1-0) right?
 
the answer I am getting (-7129.63) is apparently incorrect. can you work it out and see if I am just doing my math wrong?
 
i don't know! i am doing the equation and I am getting that answer, but the computer is saying it is the wrong answer!
 
well I'm getting a different answer than you also. Check your units. Stay consistent.